Let `f[0, 1] -> R` (the set of all real numbers be a function.Suppose the function f is twice differentiable, `f(0) = f(1) = 0`,and satisfies `f'(x) – 2f'(x) + f(x) leq e^x, x in [0, 1]`.Which of the following is true for `0 lt x lt 1 ?`

Options

A. `0ltf(x)ltoo`
B. `-1/2ltf(x)lt1/2`
C. `-1/4lt f(x)lt1`
D. `-ooltf(x)lt0`

Answer

4

Explanation

Give `f(x)-2f(x)+f(x)ge^(x)`
The terms to the L.H.S suggest htat we have to consides function `g(x)=e^(-x)f(x)`
Now `g(X)=f(x)e^(-x)-e^(-x)f(x)`
and `g(X)=f(x)e^(-x)-f(x)e^(-x)-e^(-x)f(x)+e^(-x)f(x)`
`=e^(-x)f(x)-2f(x)+f(x)`
Give that f(x)-2f(x)-2f(x)+f(x)
Given that `f(x)-2f(x)+f(x)gee^(x)`
or `g(x)=e^(-x)(fx)-2f(x)ge1ge0`
or g(X) is concave upwared and g(0)=g(1)=0
Hence `g(X)lt0forallx in (0,1) or e^(-x)f(x)lt 0` or
`f(X)lt0forall x in (0,1)`